Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Scanner;

import static java.lang.System.in;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(in);
        int n = scanner.nextInt();
        double[] arr = new double[n];
        for (int i = 0; i < n; i++) {
            arr[i] = scanner.nextDouble();
        }
        double[] dp = new double[n+1];// probability for i number of heads
        dp[0]=1.0;
        for(int coin=0;coin<n;coin++){
            for (int i = coin+1; i>=0; i--) {
                dp[i] = (i==0 ? 0 : dp[i-1]*arr[coin]) + dp[i]*(1d-arr[coin]);
            }
        }
        double res=0;
        for(int i=(n+1)/2;i<=n;i++)
            res +=dp[i];

        System.out.println(res);
    }
}